Rationality, Irrationality, and Wilf Equivalence in Generalized Factor Order

Let $P$ be a partially ordered set and consider the free monoid $P^*$ of all words over $P$. If $w,w'\in P^*$ then $w'$ is a factor of $w$ if there are words $u,v$ with $w=uw'v$. Define generalized factor order on $P^*$ by letting $u\le w$ if there is a factor $w'$ of $w$ having...

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Veröffentlicht in:The Electronic journal of combinatorics 2009-12, Vol.16 (2)
Hauptverfasser: Kitaev, Sergey, Liese, Jeffrey, Remmel, Jeffrey, Sagan, Bruce E.
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Sprache:eng
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Zusammenfassung:Let $P$ be a partially ordered set and consider the free monoid $P^*$ of all words over $P$. If $w,w'\in P^*$ then $w'$ is a factor of $w$ if there are words $u,v$ with $w=uw'v$. Define generalized factor order on $P^*$ by letting $u\le w$ if there is a factor $w'$ of $w$ having the same length as $u$ such that $u\le w'$, where the comparison of $u$ and $w'$ is done componentwise using the partial order in $P$. One obtains ordinary factor order by insisting that $u=w'$ or, equivalently, by taking $P$ to be an antichain. Given $u\in P^*$, we prove that the language ${\cal F}(u)=\{w\ :\ w\ge u\}$ is accepted by a finite state automaton. If $P$ is finite then it follows that the generating function $F(u)=\sum_{w\ge u} w$ is rational. This is an analogue of a theorem of Björner and Sagan for generalized subword order. We also consider $P={\Bbb P}$, the positive integers with the usual total order, so that $P^*$ is the set of compositions. In this case one obtains a weight generating function $F(u;t,x)$ by substituting $tx^n$ each time $n\in{\Bbb P}$ appears in $F(u)$. We show that this generating function is also rational by using the transfer-matrix method. Words $u,v$ are said to be Wilf equivalent if $F(u;t,x)=F(v;t,x)$ and we prove various Wilf equivalences combinatorially. Björner found a recursive formula for the Möbius function of ordinary factor order on $P^*$. It follows that one always has $\mu(u,w)=0,\pm1$. Using the Pumping Lemma we show that the generating function $M(u)=\sum_{w\ge u} |\mu(u,w)| w$ can be irrational.
ISSN:1077-8926
1077-8926
DOI:10.37236/88